22x+1= 8⇒ 22x+1= 23⇒ 2x + 1 = 3⇒ 2x = 2⇒ x = 1.

Solve for X: 22x+1= 8

Solve for x:
2^2x+1= 8

Sum

2^2x+1= 8
⇒ 2^2x+1= 2³
⇒ 2x + 1 = 3
⇒ 2x = 2
⇒ x = 1.

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Solving Exponential Equations

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Chapter 6: Indices - Exercise 9.1

Chapter 6 Indices
Exercise 9.1 | Q 9.01

Solve : `[3^x]^2` : 3x = 9 : 1

Solve:

2^2x + 2^x+2 − 4 × 2³ = 0

If a^x = b^y = c^z and b² = ac, prove that: y = `[2xz]/[x + z]`

Prove that : `((x^a)/(x^b))^( a + b - c ) (( x^b)/(x^c))^( b + c - a )((x^c)/(x^a))^( c + a - b)`

Solve : 3^x-1× 5^2y-3 = 225.

Evaluate the following:

`(8/27)^((-2)/3) - (1/3)^-2 - 7^0`

Find the value of k in each of the following:

`(1/3)^-4 ÷ 9^((-1)/(3)` = 3^k

If x = `3^(2/3) + 3^(1/3)`, prove that x³ - 9x - 12 = 0

If 2400 = 2^x x 3^y x 5^z, find the numerical value of x, y, z. Find the value of 2^-x x 3^y x 5^z as fraction.

Prove the following:
(x^a)^b-c x (x^b)^c-a x (x^c)^a-b= 1